Proposition (commutative ring extension is a module):
Let be a commutative ring and let be a commutative extension ring of . Then with its own addition and the restriction of the multiplication of to is a module over .
Proof: From the axioms holding for rings, we deduce the module axioms as follows as follows: Let and . Then
- (commutativity of multiplication)
the ring axiom that's being used being indicated in the brackets.
Proposition (Let be a ring extension. Then the function
defines a function from ideals of to ideals of .):
Proof: Indeed, because it is certainly closed under addition and multiplication by elements of .
Proposition (Let be a ring extension, and suppose that is a multiplicative set. Then is a multiplicative set of .):
Proof: is closed under multiplication because both and are.